 In this video, I'm going to be doing a walkthrough of a parallel RLC circuit. Now, in series circuits, it was all about the impedance triangle. In a parallel circuit, we don't want to use impedance triangle. Impedance is a bully. It can be done, but it's so hard and convoluted it is not even worth my time to talk about. What does work is power. And remember this statement, power is my friend. What we can do, and what you'll see we're going to do is we're going to build a power triangle. We're going to build a resistive element here. We'll get our inductive bars going up. We'll get our capacitive bars pushing back down on it. Then from that, we'll build an overall triangle over here and we'll figure out our VA, get our current, then get our impedance from that. So there's a bit more of a process to it than there was in a series circuit, but it's not that bad once you get the rhythm of it. So let's assign some values to this and let's get started. Okay, here's the circuit we're going to be playing with. I got five ohms of resistance. I've got 10 ohms of inductive reactants. I got 20 ohms of capacitive reactants. I got a source voltage of 600 volts at 60 Hertz. And what we're going to be trying to solve for is the impedance of the circuit and the current, the eye line of the circuit. Now, again, you could in this case figure out by building a current triangle. You could figure out the current going through here, the current going through here, the current going through here and build yourself a current triangle. Again, I don't like to do that because that only works when you have a fully resistive leg, a fully inductive leg with no resistance and a purely capacitive leg. Again, you could do that in this case, but just get in the habit of building yourself a power triangle and you'll be safe no matter what. So our first step is let's figure out what the resistive power is across this guy. I've got 600 volts from that point to that point, from that point straight to that point. I can go ahead and I can use E squared over R to get my power being dissipated through this resistor, which works out to be 72 kilowatts. Then what I can do is take the same voltage because I've got 600 volts from that point to that point. 600 squared divided by the inductive reactants of 10 ohms will give me my inductive power or my bars, my reactive power of 36 k bar. Now I've got 36 k bars heading up this way. I've got this capacitor here. I'm going to do the same thing because I've got 600 volts from there to there, which means I can go 600 squared over 20 ohms and I get 18 k bar capacitive. So I've got all my powers worked out now. I've got my kilowatts on the bottom there. I've got my k bars inductive and I've got my k bars capacitive. Let's get together and build ourselves a little or nice sized power triangle. So again, just like every power triangle we've ever built, power goes on the bottom, resistive power, 72 kilowatts. In this case here, I've got, let me just get my pen out here, I've got 36 inductive bars heading up. So I've got 36 bars heading this way. But then I've got the capacitive bars which are pushing down 18 k bar that way. So I end up with a net k bar of 36 minus 18 gives me a net of 18 k bar. Now I can use that k bar, that 18 k bar and those kilowatts to figure out what my BA is using Pythagoras because we love the triangle and I go 18 squared plus 72 squared gives me 74.2 squared over there. And that's my KBA. So let's go through that back in our circuit and see how this all plays out. So now we've got this 74.2 KBA, we've got 600 volts here. The first thing we're going to do is try to work out what our current is. We've got a source voltage of 600 volts, we've got your kilovolt amps over here, your apparent power. So I'm going to take 74,200, divide that by 600 because its power divided by voltage gives me current. In this case that works out to be 124 amps. Now we have one more step to take here, we need to figure out what the impedance of the circuit is. Well again, if we have source voltage and line current, why don't we just take the source voltage, 600 volts, divide it by 124 amps and then we will get our overall impedance of the circuit. In this case it works out to be 4.83 ohms. And there you go, we've figured it out. Again, I just caution you whenever you're dealing with a parallel circuit, use power triangles. There will always be enough information for you to have a power triangle. Do not use impedance as a bully. Do not use current, current will eventually let you down. Power is your friend.